General Physics II
General Physics II PHYS 112
Community College of Philadelphia
Popular in Course
SC Jordan Allgood
verified elite notetaker
Popular in Physics 2
This 11 page Class Notes was uploaded by Ms. Allie Wisoky on Sunday October 11, 2015. The Class Notes belongs to PHYS 112 at Community College of Philadelphia taught by David Cattell in Fall. Since its upload, it has received 62 views. For similar materials see /class/221234/phys-112-community-college-of-philadelphia in Physics 2 at Community College of Philadelphia.
Reviews for General Physics II
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 10/11/15
Alternating Current Circuits Review of rms values rms values are rootmeansquare values of quantities such as voltage and current that vary periodically with time In AC circuits voltage and current vary sinusoidally with time vVsinoat iIsinnt where Vand I are the voltage and current amplitudes respectively n is the angular frequency 03 21Tf where f is the frequency and I is a phase constant that we will discuss later The rms values of voltage and current are de ned to be Where the overbar indicates the average value of the function over one cycle Since the average value of sinze over one cycle is 12 we get V L and Inns I We 3 Note that these formulas are valid only if the voltage varies sinusoidally with time What we will study in this chapter is what happens to the current and power in an AC series circuit if a resistor a capacitor and an inductor are present in the circuit Resistors and Resistance If just a resistor of resistance R is connected across an AC generator the generator is said to have a purely resistive load The phase constant I is zero and we write v Vsinnt i Isinnt and Vms I mSR Since the angle for v and i is the same the instantaneous voltage and current are said to be in phase Note that m2 I ms is a constant independent of the frequency f of the AC generator We assume that the resistor maintains its resistance regardless of how fast or slow the generator s armature is turning R of course is measured in ohms For a purely resistive load the average power delivered to the circuit by the generator is given by 73 1V or 13 13 which are analogous to the familiar formulas for DC circuits P as usual is measured in watts Rev 12308 Capacitors and Capacitive Reactance Now let us connect just a capacitor of capacitance C across an AC generator In this case the generator is said n to have a purely capacitive load The phase constant 4 1s 3 and we wr1te rms vVsinoat i1sin mtgj and V Im SXC where X c is called the capacitive reactance Capacitive reactance like resistance is measured in ohms Since the angle for the instantaneous current is greater than the angle for the instantaneous voltage by n2 radians or 90 the current is said to lead the voltage by 90 or lead the voltage by a quarter cycle Remember that a full cycle is 360 a complete trip around a circle Vms For a capacitive load the ratio is not a constant independent of the frequency of the generator It can be rms shown that in fact V 1 sothat XC71 Inns anC anC l l l V Un1ts check 7 7 7 ohms See F1gure 232 on page 727 ofyour text HZ F L Q g L A S V S V For a purely capacitive load the average power delivered to the circuit by the generator is zero The reason for this is that the instantaneous voltage and current in the circuit are exactly 90 out of phase Over one cycle the generator delivers as much power to the capacitor as it gets back from the capacitor Remember that over a generator cycle the capacitor will charge then discharge Example Two stripped wires from the end of a lamp cord are soldered to the terminals of a 200 HF capacitor The lamp cord which has a standard electric plug on the other end is then plugged into a 120 V 60 Hz AC outlet Do not try this at home a Find the reactance of the capacitor 1 1 X 7 X X 133 hm C 21th C 2n602x10quot C 0 S b Find the rms current drawn from the wall outlet Vms1msXc Lush 1ms120V39 X 133 o Inductors and Inductive Reactance Now let us connect just an inductor of inductance L across an AC generator In this case the generator is said to n have a purely inductive load The phase constant 4 1s 5 and we wr1te vVsinoat i1sin mt g and VmsImsXL where X L is called the inductive reactance Inductive reactance like resistance is measured in ohms Since the angle for the instantaneous current is smaller than the angle for the instantaneous voltage by n2 radians or 90 the current is said to lag the voltage by 90 or lag the voltage by a quarter cycle Alternatively one can say that the voltage leads the current by 90 or the voltage leads the current by a quarter cycle V For an 1nduct1ve load the ratlo quotquot5 1s not a constant independent of the frequency of the generator It can be shown that in fact Vrms 21th so that XL anL rms Units check Hz H A ohms See Figure 236 on page 729 ofyour text s For a purely inductive load the average power delivered to the circuit by the generator is zero The reason for this is that the instantaneous voltage and current in the circuit are exactly 90 out of phase Over one cycle the generator delivers as much power to the inductor as it gets back from the inductor Remember that over a generator cycle the induced emf in the inductor will reverse direction Example Two stripped wires from the end of a lamp cord are soldered to the terminals of a 200 mH inductor The lamp cord which has a standard electric plug on the other end is then plugged into a 120 V 60 Hz AC outlet Do not try this at home a Find the reactance of the inductor XL anL XL 2n600200 b Find the rms current drawn from the wall outlet V I 120v rms XL rms Q5 rms RCL Series Circuits n itr ninrlutr nilquot An i Y See the figure e rnatliernaueal analysis othls eireuitrequires the solution However tliere is a E i l t o p sor ramtes counterclockwise about the origin vvitli angular o the angular frequency ofthe Ac generator The E51 i ysinm phasor represents eitlier voltage or eurrent and its yr eornponent is the instantaneous value ofthe quantity it represents vvill assurne tlnat at any instant the eurrenttlirougli each eireuit 3 Y We elernent is giv n b 15mm The eurrentpliasorlias length 1 and makes an angle of mt r b vvitln respeetto tlie inanis At any instantits yrcomponent equals the eunentin tlie eireuit ow eonsiolertlie voltage phasor othe resistor The instantaneous voltage aeross the resistor isiust iR 1R5m0t b or vR VR sm0t b The lengtli ofthe resistor39s voltage phasons the voltage arnplituole K At any instant the angle itrnakes vvitln tlie inanis is out 7 b The r eornponent ofthls phasor is then szm t Qv ss the resistor Note that the li is the instantaneous voltage aero Henee the voltage nie eunent an voltage aeross the resistor are mphase phasor forthe resistor lies on top the eurrent phasor Now eonsioler the voltage phasor for the critically xmpmmnt to remember the phase eunent and voltage for a eapaeitor Does the apaclm Here itis relationship between the eurrent leaol or lag the voltage by 90 sinee tlie phasors rotate eountereloekvvise the voltage phasor forthe eapaeitor must lie 90 clackwxse from the eunent phasor Now consider the voltage phasor for the inductor It is critically important to remember the phase relationship between the current and voltage for an inductor Does the current lead or lag the voltage in an inductor By how many degrees The current lags the voltage by 90 Since the phasors rotate counterclockwise the voltage phasor for the inductor must lie 90 counterclockwise from the current phasor Note that the voltage phasors for the inductor and the capacitor lie along the same line We have arbitrarily assumed that VL is larger than Vc Using the rules of vector addition we may combine them to obtain the next diagram By Kirchhoff s loop rule the voltage drops across the capacitor resistor and inductor must at any instant equal the voltage rise across the generator This will be satis ed if the vector sum of the VL Vc and the VR phasors matches the voltage phasor of the generator See the last diagram below From the last diagram we obtain some very important relationships In particular note that V2VL VC2VRZ or V IL IC2VRZ since VLIXL VCIXC and VRIR we canwrite V IXL IXC2IRZ or VI XL XCZRZ or VIZ where Z XL XC2RZ Z is called the impedance of the circuit and is measured in ohms Note that we have dropped the rms subscripts for the voltage and the current in the V IX formulas above because the formulas are also valid if we replace each rms value with its corresponding amplitude the square root of 2 cancels from both sides of each equation We can now find a formula for the phase I of the current From the right triangle with sides V VR and VL Vc in the diagram above we have tand so that tan VR IR Average Power On average only the resistance in the RCL series circuit consumes power The average rate of power consumption is given by 7 2 P ImsR Z XL Xc The triangle at the right is useful to remember since one can quickly obtain the formulas that were derived above from it R Z XL XC2RZ and tan 7XLIEXC also note that g coscl so that R Z coscl and 13 linsZ coscl or 13 Ims1msZcos from which we obtain V O 0 V 9 ms nu coscl is called the power factor of the RCL circuit XL XC Using the formula tand we make the following observations and definitions If XL gt Xc I gt 0 and the circuit is said to have an inductive load If X L lt X C I lt 0 and the circuit is said to have a capacitive load If XL Xc I 0 and the circuit is said to have a resistive load Example A series RCL circuit has a 750 Q resistor a 200 uF capacitor and a 550 mH inductor connected across an 800 volt rms AC generator operating at 128 Hz a Is the load on the circuit inductive capacitive or resistive What is the phase angle 1 XL anL XL 2nl2855 X104 442 Q 1 1 X 7 X C 2nfC C 211l2820gtlt10395 Since XC gt XL the load is The phase angle is ltgt arctan b arctan 612 Q R 750 I 0223 rad or b 0 3 1 D quot1 What is the rms current in the circuit To answer this question we must determine the circuit s impedance Z then use I rm VrmSZ Z lXL Xc2 R2 Z 442 6122 7502 Z 769 Q In 1m 800 V In 104 A Z 769 Q Write the formula for the current in the circuit as a function of time i Isin0at 1 where I is the current amplitude I Ims I 1041414 I 147 A 03 21tf 03 2n128 03 804 rads i l47sin804t0223 t in seconds and i in amperes Note the use of radians Find the rms voltage across each circuit element Vms ImsR Vm 104 A750 Q Vms 780 V chs ImsXc Vm 104 A612 Q VCms 636 V Vms ImsXL Vms 104 A442 Q Vms 460 V Question Shouldn t these Voltages add to 800 V Answer No One must take into account the phase of the Voltage across each element See part e Find the instantaneous Voltage across each circuit element at t 0 seconds The voltage across the capacitor VR iR VR 147750sin0223 VR 244 V lags the current by 90 vc tnZXC vc 147612sin0223 157 vc 877 V The voltage across the inductor leads the current by 90 vL 1MXL vL 147442sin0223157 vL 633 V Question Why do these Voltages add to zero Answer Their sum is in agreement with Kirchhoff s loop rule the Voltage across the generator is v Vsin0at or v 8005sinlt804tgt 0 at t 0 s Find the aVerage power deliVered to the circuit by the generator 13 I V cos 13 104 A800 Vcos 0223 m1 m1 F811kW The Limiting Behavior of Capacitors and Inductors Unlike a resistor which has a constant 39 R 39 J r J of the ac q quot and inductors have reactances that do depend on it The inductive reactance is given by X L anL If f is large so is X L and the inductor acts almost like an open circuit If f is small so is X L and the inductor acts almost like a short circuit V R Vgeff Vin L Vom Frequency f This circuit can be regarded as a highpass lter At veryhigh frequencies the inductor has a high reactance and acts almost like an open circuit Thus the current is low the voltage drop in the resistor is low and Vout Vin At verylow frequencies the inductor has a low reactance and acts like a short circuit The output voltage is Virtually zero Hence the circuit passes highfrequency AC voltages but stops lowfrequency AC voltages The capacitive reactance is given by l anC C If f is large X c is small and the capacitor acts almost like a short circuit If f is small X c is large and the capacitor acts almost like an open circuit Vgeff VI c Vout 111 Frequency f l n r e quot msye m e w 39 waxm Vm Theoutput voltageisvirtuallyzero quot 39 39 1 a A Voltages Example stereospeakerA 2 quot tweeter quot 39 39 speakerra woofer 1 39 A 39 I e u r 39 Vmisthevoltage Wm r Vamid 39 39 If RisSOohmsc L 39 39 39 39 atoutput amp 5 7X6 W Vm X R2 1 2 l gR2XC e Squarebothsides 2 JXCR 2 1 2 2 2 1 2 7 X R X 7R A c J c 4 30300 oz Xc173 O Xc 2qu 1 8000Hz 7 f 27t8000173 Remark The free quot 39 39 frequency Inme above example 3000 Hzis me crossover frequency Example frequency of a 1000000 Hz a For ahigh frequency me inductors actlike open circuits andthe capacitor acts like a short circuit effectively producing me circuit wnin me diagram on the nextpage R1 The impedance is now just the net resistance of the circuit m Since the resistors are in series RR1R2 R1k2k R3kQ Z3kQ b For a low frequency the inductors act as short circuits and the capacitor acts as an open circuit effectively producing the circuit shown in the diagram below The impedance is now just the net resistance of the circuit Since the resistors are in parallel iii i1 R067kQ R2 R lk 2k R 2k Z 067 kQ 2k As the frequency of the AC generator is changed from very low values to very high values the impedance of the circuit h39 l l l iii R R1 R2 will increase from the lower limit of 067 kQ to the upper limit of 3 k9 Note The formula for impedance we found earlier Z XL X C 2 R2 does not apply to the given circuit in this example because the circuit elements are not connected in series The formulas for the reactances however always apply Electrical Resonance For an RCL series circuit the current amplitude is given by 1K V Z XL Xc2 R2 where Vis the voltage amplitude If V R C and L are fixed and the frequency of the AC generator is variable we can change the reactances of the inductor and capacitor by changing the frequency of the generator As the frequency of the generator changes so does the impedance Z of the circuit and the current amplitude I If we look at the above formula we see that Z can be minimized made as small as possible by making the reactances X L and X C equal to one another The current amplitude I will then be maximized made as large as possible If these conditions are met electrical resonance is said to occur in the circuit The RCL series circuit is said to be at resonance For resonance This value of f is called the resonant frequency of the RCL series circuit At resonance the phase angle 4 is zero and the circuit has a resistive load The power factor coscl is l and maximum power is delivered to the circuit by the generator At resonance the impedance Z equals the resistance R Example An RCL series circuit is powered by an AC generator with rrns voltage 200 V R 200 Q C 500 HF L 200 mH a Find the resonant frequency of the circuit b C f 1 f 1 f 1 2mLC 2n 0200 H500gtlt10 6 2nJ100xl0 6 f 1030C21000 f159HZ dc52 TVS Find the rms current at resonance V I 200v ms I 100A 2009 rms Find the average power delivered to the circuit at resonance 13 100 A200 v 13 V rms rms
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'